If C(16,r) = C(16,r+2), then find r. Explain how you know.
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Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
add a comment |
up vote
2
down vote
favorite
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?
combinations
combinations
asked 2 hours ago
Gold Pony Boy
152
152
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3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
add a comment |
up vote
3
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
add a comment |
up vote
4
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
Note that $binom{n}{k} = binom{n}{n-k}.$
What equation does this give for $r$ in your scenario? What is the solution to that equation?
Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.
edited 5 mins ago
answered 2 hours ago
Display name
789313
789313
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
add a comment |
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
Your equation does not imply $$binom na=binom nbimplies a=n-b$$
– Kemono Chen
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
@KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
– Display name
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
– Kemono Chen
1 hour ago
add a comment |
up vote
3
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
add a comment |
up vote
3
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?
answered 2 hours ago
Carl Schildkraut
10.6k11438
10.6k11438
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
add a comment |
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
Okay, I see the connection now, so r must be 7...?
– Gold Pony Boy
1 hour ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
@GoldPonyBoy Yes.
– Carl Schildkraut
34 mins ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
add a comment |
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
add a comment |
up vote
0
down vote
up vote
0
down vote
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
$C_r^{16}=C_{r+2}^{16}$
Then using the definition of combination you'll get
$$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
$$implies frac{1}{(16-r)!}=frac{1}{(r+2)(r+1).(14-r)!}$$
$$implies (16-r)(15-r)=(r+2)(r+1)$$
$$implies 240-31r+r^2=r^2+3r+2$$
$$implies 34r=238$$
$$implies r=7$$
edited 1 hour ago
answered 1 hour ago
Fareed AF
36711
36711
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
add a comment |
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
– Gold Pony Boy
1 hour ago
2
2
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
– mathematics2x2life
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
@mathematics2x2life Actually wait,where did the r! from the left side go after the first step?
– Gold Pony Boy
1 hour ago
1
1
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
@Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
– Fareed AF
1 hour ago
add a comment |
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